Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}9x+4y &= 1 \\ 6x+y &= 1\end{align*}$
Explanation: Begin by moving the $y$ -term in the second equation to the right side of the equation. $6x = -y+1$ Divide both sides by $6$ to isolate $x$ $x = {-\dfrac{1}{6}y + \dfrac{1}{6}}$ Substitute this expression for $x$ in the first equation. $9({-\dfrac{1}{6}y + \dfrac{1}{6}}) + 4y = 1$ $-\dfrac{3}{2}y + \dfrac{3}{2} + 4y = 1$ Simplify by combining terms, then solve for $y$ $\dfrac{5}{2}y + \dfrac{3}{2} = 1$ $\dfrac{5}{2}y = -\dfrac{1}{2}$ $y = -\dfrac{1}{5}$ Substitute $-\dfrac{1}{5}$ for $y$ in the top equation. $9x+4( -\dfrac{1}{5}) = 1$ $9x-\dfrac{4}{5} = 1$ $9x = \dfrac{9}{5}$ $x = \dfrac{1}{5}$ The solution is $\enspace x = \dfrac{1}{5}, \enspace y = -\dfrac{1}{5}$.